Question: $70\%$ of a certain species of tomato live after transplanting from pot to garden. Najib transplants $3$ of these tomato plants. Assume that the plants live independently of each other. Let $X =$ the number of tomato plants that live. What is the probability that exactly $2$ of the $3$ tomato plants live? You may round your answer to the nearest hundredth. $P(X=2)=$
Solution: Without a fancy calculator Having $2$ plants live from the $3$ transplanted means Najib needs to have $2$ plants live and $1$ plant not live. We know $P({\text{live}})={70\%}$ and $P({\text{not}})={30\%}$. Since we are assuming independence, let's multiply probabilities to find the probability of $2$ plants living followed by $1$ plant not living: $P({\text{LL}}{\text{N}})=({0.7})^2({0.3})=0.147$ This isn't our final answer, because there are other ways for $2$ of $3$ plants to live after transplant (for example, LNL). How many different ways are there? We can use the combination formula to find how many ways there are for $2$ of $3$ plants to live after transplant: $\begin{aligned} _n\text{C}_k&=\dfrac{n!}{(n-k)!\cdot k!} \\\\ _3\text{C}_2&=\dfrac{3!}{(3-2)!\cdot2!} \\\\ &=\dfrac{3 \cdot \cancel{2 \cdot 1}}{(1) \cdot \cancel{2 \cdot 1}} \\\\ &=3 \end{aligned}$ There are $3$ ways for $2$ of $3$ plants to live after transplant. Do they all have the same probability? Each of the $3$ ways has the same probability that we already found: $\begin{aligned} P({\text{LL}}{\text{N}})&=({0.7})^2({0.3})=0.147 \\\\ P({\text{L}}{\text{N}}{\text{L}})&=({0.7})^2({0.3})=0.147 \\\\ P({\text{N}}{\text{LL}})&=({0.7})^2({0.3})=0.147 \end{aligned}$ So we can multiply this probability by $3$ since that is how many ways there are for $2$ of $3$ plants to live after transplant: $\begin{aligned} P(X=2)&=3(0.7)^2(0.3) \\\\ &=3(0.147) \\\\ &=0.441 \end{aligned}$ Answer $P(X=2)=0.441\approx0.44$